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Sầm Sơn là khu nghỉ mát nổi tiếng của tỉnh Thanh Hóa. Từ bờ biển nhìn ra, khu bãi tắm giống như một đường cong xanh mềm mại. Lại gần thấy nước biển rất xanh và sạch. Sóng biển hiền hòa vỗ vào bờ như ru ngủ những hàng dừa. Tắm biển Sầm Sơn điều thú vị nhất là được những con sóng mạnh mẽ, trong lành đẩy lên rồi hạ xuống như đùa giỡn với ta. Nếu muốn được ngắm nhìn toàn cảnh bờ biển, ta có thể đi ca nô ra xa bờ một chút. Ca nô rẽ nước tạo thành những vệt trắng như tuyết. Trên ca nô, ta được ngắm nhìn một bên là bờ biển đông vui nhộn nhịp một bên là nước biển mênh mông xanh trong rất tuyệt vời. Mấy ngày ở Sầm Sơn, được thưởng thức vẻ đẹp của cảnh đẹp kì thú này càng khiến em thấy tự hào về đất nước Việt Nam giàu đẹp, trù phú của mình. >> Tham khảo: Các bà
Câu 1: Explain the difference between connectionless unacknowledged service and connectionless acknowledged service How the protocols that provide these services differ? Connectionless service is a network communication service that does not establish a dedicated communication path between the sender and receiver In connectionless unacknowledged service, also known as "fire and forget," the sender sends a message without ensuring that it has been received by the receiver The receiver may or may not receive the message, and there is no confirmation sent back to the sender In contrast, in connectionless acknowledged service, the sender transmits a message and waits for a confirmation of receipt from the receiver If the receiver receives the message, it sends an acknowledgement back to the sender If the sender does not receive an acknowledgement, it can assume that the message was not received The protocols that provide these services differ in their handling of packet loss and protocol overhead Connectionless unacknowledged service protocols, such as UDP (User Datagram Protocol), are lightweight and have minimal overhead but provide no guarantee of message delivery Connectionless acknowledged service protocols, such as ICMP (Internet Control Message Protocol) and ARP (Address Resolution Protocol), have more overhead but provide greater reliability by sending acknowledgements of successful message delivery Câu Explain the difference between connection-oriented acknowledged service and connectionless acknowledged service How the protocols that provide these services differ? Connection-oriented acknowledged service is a network communication service that establishes a dedicated connection between the sender and receiver before any data transmission takes place In this service, the receiver sends an acknowledgement to the sender after receiving each packet of data In contrast, connectionless acknowledged service does not establish a dedicated connection before data transmission The sender simply sends packets to the receiver, and the receiver sends acknowledgements back to the sender for each received packet The protocols that provide these services differ in their handling of packet loss and protocol overhead Connection-oriented acknowledged service protocols, such as TCP (Transmission Control Protocol), use a three-way handshake to establish a reliable connection between the sender and receiver This protocol also implements flow control and congestion control mechanisms to ensure efficient data transfer and minimize packet loss Once the data transmission is complete, the connection is terminated Connectionless acknowledged service protocols, such as UDP (User Datagram Protocol), not establish a dedicated connection with the receiver Instead, they simply send the packets to the receiver without guaranteeing their delivery or order These protocols have lower overhead and are useful in applications where speed is more important than accuracy, such as online gaming and live streaming In summary, connection-oriented acknowledged service is reliable but has higher overhead and may be slower, while connectionless acknowledged service is faster but less reliable and does not guarantee the order of packet delivery Câu 3: Explain the differences between PPP and HDLC PPP (Point-to-Point Protocol) and HDLC (High-level Data Link Control) are both protocols used for data link layer communication While they share many similarities, there are several differences between the two protocols Flexibility: PPP is a more flexible protocol than HDLC It can be used to carry multiple network layer protocols, including IP, IPX, and AppleTalk, whereas HDLC is primarily designed for carrying only one protocol Error Detection: PPP has a better error detection mechanism as compared to HDLC PPP uses a cyclic redundancy check (CRC) for detecting errors, while HDLC uses a frame check sequence (FCS) The CRC is considered more effective in detecting errors in data transmission Configuration: PPP is easier to configure than HDLC PPP uses a configuration protocol called LCP (Link Control Protocol), which automates the configuration process In contrast, HDLC requires manual configuration of parameters, such as the address field and control field Authentication: PPP supports authentication mechanisms such as PAP (Password Authentication Protocol) and CHAP (Challenge Handshake Authentication Protocol), which authenticate the identity of the sender HDLC does not support any authentication mechanism Sliding Window Protocol: HDLC uses a sliding window protocol for flow control, while PPP does not The sliding window protocol is used to optimize the flow of data between the sender and receiver, ensuring that the receiver is not overwhelmed with too much data at once In summary, PPP is a more flexible, easier to configure and better error detection mechanism than HDLC Additionally, PPP offers mechanisms like authentication, which HDLC lacks However, HDLC uses a sliding window protocol for flow control, which PPP does not provide Câu 4: A 1.5 Mbps communications link is to use HDLC to transmit information to the moon What is the smallest possible frame size that allows continuous transmission? The distance between earth and the moon is approximately 375,000 km, and the speed of light is x 108 meters/second To determine the smallest possible frame size that allows continuous transmission, we need to calculate the round-trip time (RTT) for a signal to travel from Earth to the Moon and back RTT = * distance / speed of light Converting the distance between Earth and the Moon to meters: 375,000 km * 1000 m/km = 375,000,000 meters Plugging in the values: RTT = * 375,000,000 / x 10^8 = seconds To achieve continuous transmission on a 1.5 Mbps link with HDLC, we need to calculate the minimum frame size that can be transmitted within this RTT Minimum frame size = (Link speed * RTT) / Efficiency The efficiency factor takes into account the protocol overhead, such as header and trailer bits Assuming an efficiency factor of 80%, we get: Minimum frame size = (1.5 Mbps * sec) / 0.8 = 11.25 megabits To convert to bytes: 11.25 megabits / bits/byte = 1.41 megabytes Therefore, the smallest possible frame size that allows continuous transmission on a 1.5 Mbps HDLC link to the Moon is approximately 1.41 megabytes Câu 5: Suppose HDLC is used over a 1.5 Mbps geostationary satellite link Suppose that 250-byte frames are used in the data link control What is the maximum rate at which information can be transmitted over the link? To determine the maximum rate at which information can be transmitted over a 1.5 Mbps geostationary satellite link using HDLC with 250-byte frames, we need to take into account the overhead of the HDLC protocol The HDLC frame consists of several fields, including flag characters, address, control, data, and CRC (Cyclic Redundancy Check) The flag characters mark the beginning and end of the frame and are each one byte long The address and control fields are each one byte long The CRC field is two bytes long Therefore, the total size of an HDLC frame is: Frame size = Data + Address (1 byte) + Control (1 byte) + CRC (2 bytes) + Flag (1 byte) + Flag (1 byte) = Data + bytes For 250-byte frames, the total frame size is: Frame size = 250 + = 256 bytes To calculate the maximum rate of information transmission, we need to divide the link speed by the time it takes to transmit one frame, including the overhead Time to transmit one frame = Frame size / Link speed = 256 bytes / 1.5 Mbps = 0.001707 seconds Therefore, the maximum rate of information transmission is: Maximum rate = Data rate / Efficiency The efficiency factor takes into account the protocol overhead, such as the header and trailer bits Assuming an efficiency factor of 80%, we get: Maximum rate = (250 bytes * bits/byte) / 0.001707 seconds / 0.8 = 23,154 bits per second Therefore, the maximum rate at which information can be transmitted over a 1.5 Mbps geostationary satellite link using HDLC with 250-byte frames is approximately 23,154 bits per second Câu 6: Suppose that a multiplexer receives constant-length packet from N = 60 data sources Each data source has a probability p = 0.1 of having a packet in a given T-second period Suppose that the multiplexer has one line in which it can transmit eight packets every T seconds It also has a second line where it directs any packets that cannot be transmitted in the first line in a T-second period Find the average number of packets that are transmitted on the first line and the average number of packets that are transmitted in the second line Given: N = 60 data sources Probability of having a packet in a given T-second period, p = 0.1 The multiplexer can transmit eight packets every T seconds Any packets that cannot be transmitted in the first line are directed to the second line To find: Average number of packets that are transmitted on the first line Average number of packets that are transmitted in the second line We can model this scenario as a binomial distribution problem, where each data source has a probability p of transmitting a packet and there are N such sources The probability of k sources transmitting a packet is given by the binomial distribution formula: P(k) = (N choose k) * p^k * (1-p)^(N-k) where (N choose k) is the binomial coefficient, given by: (N choose k) = N! / (k! * (N-k)!) We want to find the average number of packets that are transmitted on the first line and the second line Let X be the total number of packets generated in a T-second period Then, we can divide X into two parts: Y, the number of packets transmitted on the first line, and Z, the number of packets transmitted on the second line Since we can transmit up to eight packets on the first line, we have: Y = min(X, 8) For the remaining packets, we send them on the second line, so we have: Z = max(0, X - 8) The expected value of Y is given by: E(Y) = Σ[k=0 to 8] k * P(k)k=0 to 8] k * P(k) where P(k) is the probability of k sources transmitting a packet, as computed using the binomial distribution formula Similarly, the expected value of Z is given by: E(Z) = Σ[k=0 to 8] k * P(k)k=9 to N] k * P(k) where P(k) is again the probability of k sources transmitting a packet Let's compute these values using the given parameters: N = 60 p = 0.1 T = second (since we are considering a T-second period) M = (maximum packets that can be transmitted on the first line) First, let's calculate the probabilities of k sources transmitting a packet: P(k) = (N choose k) * p^k * (1-p)^(N-k) For k = to 8: P(0) = (60 choose 0) * 0.1^0 * 0.9^60 = 0.026 P(1) = (60 choose 1) * 0.1^1 * 0.9^59 = 0.157 P(2) = (60 choose 2) * 0.1^2 * 0.9^58 = 0.318 P(3) = (60 choose 3) * 0.1^3 * 0.9^57 = 0.306 P(4) = (60 choose 4) * 0.1^4 * 0.9^56 = 0.185 P(5) = (60 choose 5) * 0.1^5 * 0.9^55 = 0.080 P(6) = (60 choose 6) * 0.1^6 * 0.9^54 = 0.027 P(7) = (60 choose 7) * 0.1^7 * 0.9^53 = 0.007 P(8) = (60 choose 8) * 0.1^8 * 0.9^52 = 0.001 For k = to 60: P(k) = (N choose k) * p^k * (1-p)^(N-k) For simplicity, we can use the complement rule and subtract the sum of probabilities from to from 1: P(k) = - Σ[k=0 to 8] k * P(k)k=0 to 8] P(k) = - (P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8)) = 0.996 Now, let's calculate the expected values of Y and Z: E(Y) = Σ[k=0 to 8] k * P(k)k=0 to 8] k * P(k) = 00.026 + 10.157 + 20.318 + 30.306 + 40.185 + 50.080 + 6*0.027 + Câu 7: Consider the transfer of a single real-time telephone voice signal across a packet network Suppose that each voice sample should not be delayed by more than 20 ms a Discuss which of the following adaptation functions are relevant to meeting the requirements of this transfer: handling of arbitrary message size; reliability and sequencing; pacing and flow control; timing; addressing; and privacy, integrity and authentication b Compare a hop-by-hop approach to an end-to-end approach to meeting the requirements of the voice signal a To meet the requirement of transferring a single real-time telephone voice signal across a packet network with a maximum delay of 20 ms, the following adaptation functions are relevant: Timing: The timing adaptation function is critical in ensuring that each voice sample is delivered within the required deadline The network must be able to synchronize its clock with the sender and receiver to maintain the required time intervals between packets Reliability and sequencing: To ensure that each voice sample is delivered without loss or misordering, the reliability and sequencing adaptation function is necessary This requires the use of error detection and correction mechanisms, as well as sequencing and resequencing of packets at the receiver end Pacing and flow control: To prevent packet loss due to congestion, pacing and flow control mechanisms are necessary These mechanisms regulate the rate at which packets are transmitted and received to match the capacity of the network Addressing: Addressing is necessary to identify the source and destination of each voice sample It also enables routing of packets through the network b There are two approaches for meeting the requirements of a real-time telephone voice signal over a packet network: the hop-by-hop approach and the end-to-end approach The hop-by-hop approach involves implementing the required adaptation functions at each intermediate node in the packet network Each node processes the packets it receives before forwarding them to the next node This approach can introduce additional delays and overhead due to processing at each node Furthermore, if a node fails, the entire communication may become compromised The end-to-end approach involves implementing the required adaptation functions only at the endpoints of the communication path, i.e., the sender and receiver of the voice signal The packets are transmitted through the network without modification, and any required processing is performed at the endpoints This approach minimizes delays and overhead, but it may not be suitable for networks with high packet loss rates or variable delays In general, the end-to-end approach is preferred for real-time voice communications over packet networks because it minimizes delays and overhead However, the hop-by-hop approach may be necessary in some situations, such as when the network has high delay or loss rates, or when additional processing is necessary at intermediate nodes Câu 8: Consider the Stop-and-Wait protocol as described Suppose that the protocol is modified so that each time a frame is found in error at either the sender or receiver, the last transmitted frame is immediately resent a Show that the protocol still operates correctly b Does the state transition diagram need to be modified to describe the new operation? c What is the main effect of introducing the immediate-retransmission feature? a The modified Stop-and-Wait protocol still operates correctly because it ensures that every frame is received correctly before the next one is sent If a frame is found in error, the sender immediately resends the last transmitted frame, which guarantees that the receiver will receive a correct copy of the frame b The state transition diagram would need to be modified to reflect the new operation Specifically, a new transition would need to be added from the "Frame Received, ACK/NAK Lost" state back to the "Frame Sent" state, indicating that the sender should immediately resend the last transmitted frame in response to the error c The main effect of introducing the immediate-retransmission feature is to improve the protocol's error recovery capabilities With this feature, errors can be quickly corrected by resending the last transmitted frame This reduces the time required for error recovery and increases the overall efficiency of the protocol However, it also introduces additional network traffic, which could potentially increase congestion and delay